Question

请大神帮我看下画红圈的两道题，要过程，谢了。

Answer 1

<1> (2)
y' = (x/y)^2
y^2 dy = x^2 dx
y^3 = x^3 +C
(3) <2>
dy/dx - 3xy =x
dy/dx = x(1+3y)
∫dy/(1+3y)= ∫x dx
(1/3)ln|1+3y| + C' = (1/2)x^2
ln|1+3y| = (3/2)x^2 - 3C'
1+3y = C.e^[(3/2)x^2]
y =(1/3) { 1-C.e^[(3/2)x^2] }

追问

我向你提了一个问，里面有详细内容，请你帮我看看好吗，就是3题第2问的

我向你提了一个问，里面有详细内容，请你帮我看看好吗，就是3题第2问的

追答

就是3题第2问的

(3)
dy/dx - 3xy =x
dy/dx = x(1+3y)
∫dy/(1+3y)= ∫x dx
(1/3)ln|1+3y| + C' = (1/2)x^2
ln|1+3y| = (3/2)x^2 - 3C'
1+3y = C.e^[(3/2)x^2]
y =(1/3) { 1-C.e^[(3/2)x^2] }